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Ta có: \(\dfrac{B}{A}=\dfrac{\dfrac{1}{2016}+\dfrac{2}{2015}+\dfrac{3}{2014}+...+\dfrac{2015}{2}+\dfrac{2016}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{1+\left(1+\dfrac{2015}{2}\right)+\left(1+\dfrac{2014}{3}\right)+...+\left(1+\dfrac{2}{2015}\right)+\left(1+\dfrac{1}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=2017\)
A=3/1+3/3+3/6+...+3/2033136
A=2(3/2+3/6+3/12+...+3/4066272)
A=2.3.(1/1.2+1/2.3+...+1/2016.2017)
A=6.(1-1/2+1/2-1/3+...+1/2016-1/2017)
A=6.(1-1/2017)=336/2017
Vậy A=336/2017
\(A=\left(1+2\right).\frac{1}{2}+\left(1+2+3\right).\frac{1}{3}+...+\left(1+2+3+...+2016\right).\frac{1}{2016}\)
\(A=\left(1+2\right).2:2.\frac{1}{2}+\left(1+3\right).3:2.\frac{1}{3}+...+\left(1+2016\right).2016:2.\frac{1}{2016}\)
\(A=3:2+4:2+...+2017:2\)
\(A=3.\frac{1}{2}+4.\frac{1}{2}+...+2017.\frac{1}{2}\)
\(A=\frac{1}{2}.\left(3+4+...+2017\right)\)
\(A=\frac{1}{2}.\left(3+2017\right).2015:2\)
\(A=\frac{1}{2}.2020.2015.\frac{1}{2}\)
\(A=505.2015=1017575\)